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Is arc connected-ness well-behaved with respect to products?

That is -

$\prod X_\alpha$ is arc connected iff $X_\alpha$ is arc connected $\forall \alpha$

In this question on MathStackexchange, an answer is provided only for the reverse implication, that is -

If $X_\alpha$ is arc connected $\forall \alpha$, then $\prod X_\alpha$ is arc connected

However, I wasn't able to get an answer for the forward implication, nor have I been able to find it in any book or from Googling. So, is the forward implication true, or is there a counterexample for the same?

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  • $\begingroup$ They are equivalent when $X_\alpha$ is Hausdorff because in this context path-connected implies arc-connected. I will now assume that you are asking about T$_1$ spaces. $\endgroup$
    – D.S. Lipham
    Commented Aug 8, 2020 at 19:00
  • $\begingroup$ In fact, I'm asking for general $X_\alpha$, which need not even be $T_0$. $\endgroup$
    – Ishan Deo
    Commented Aug 8, 2020 at 19:01

1 Answer 1

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This is false when the spaces are not Hausdorff. Let $X$ be the line with two origins $\{O_1,O_2\}$ and $Y$ be the usual line. Then $X\times Y$ is arc connected because you can pick an arc that starts at $(O_1,y_1)$ travels outside $\{O_1,O_2\}\times Y$ and then comes back to $(O_2,y_2)$, but $X$ itself is not arc connected.

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  • $\begingroup$ Sorry it's early in the morning and I might be missing something obvious, but writing $X=\mathbb{R}\sqcup\mathbb{R}/\sim$ why can't I take the path $[0,\frac{1}{2}]\to[0,\frac{1}{2}]$ on the first copy of $\mathbb{R}$ and $[\frac{1}{2},1]\to[\frac{1}{2},0]$ on the second copy to link the two origins? $\endgroup$
    – Daniel Robert-Nicoud
    Commented Aug 9, 2020 at 6:17
  • $\begingroup$ @DanielRobert-Nicoud that shows that it is path connected, which is different from arc connected. You would need your map from $[0,1]\to X$ to be injective. $\endgroup$
    – Gjergji Zaimi
    Commented Aug 9, 2020 at 6:21
  • $\begingroup$ Ah I see! Thanks, I had never come across this notion tbh. $\endgroup$
    – Daniel Robert-Nicoud
    Commented Aug 9, 2020 at 6:24
  • $\begingroup$ Also since the projection of a path is still a path, the implication of the OP is true for path connected.. $\endgroup$
    – Daniel Robert-Nicoud
    Commented Aug 9, 2020 at 6:25

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